if √x=y+x find dy/dx
Answers
Answer:
Step-by-step explanation:
d(xy)=d(ey⋅log(x))=ey⋅log(x)d(y⋅log(x))=(xy)(dy⋅log(x)+y⋅d(log(x))=(xy)(log(x)dy+y/xdx)
Symmetrically : d(yx)=(yx)(log(y)dx+x/ydy).
Since d(xy)=d(yx), and simplifying by xy=yx, we get
log(y)dx+x/ydy=log(x)dy+y/xdx.
Removing the denominators leads to:
xylog(y)dx+x2dy=xylog(x)dy+y2dx
Which can be rearranged in
(xylog(y)−y2)dx=(xylog(x)−x2)dy
And finally
dy/dx=xylog(y)−y2xy⋅log(x)−x2
Note that the use of differentials (rather than derivative) allows to take full advantages of the symmetry of the equation.
9.1k views · View 8 Upvoters
Ved Prakash Sharma
Ved Prakash Sharma, former Lecturer at Sbm Inter College, Rishikesh (1971-2007)
Answered Jan 11, 2018 · Author has 7k answers and 2.9m answer views
x^y=y^x , taking log of both sides
y.log x=x.log y
y.1/x+dy/dx.log x=x.1/y.dy/dx+1.log y
dy/dx.log x-x/y.dy/dx=log y-y/x
dy/dx(log x-x/y)=(log y-y/x)
dy/dx=(log y-y/x)/(log x-x/y) , Answer
2.2k views · View 3 Upvoters
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Nikos Mantzakouras
Nikos Mantzakouras, Μ.Sc Applied Mathematics, Engineering, and Physics & Special Relativity, National Kapodistrian University ...
Answered Aug 21, 2019 · Author has 966 answers and 196.4k answer views
Originally Answered: What is the differential of xy=yxas x ?
We solving the equation…
xy=yx=>
ylogx=xlogy=>
1/xlogx=1/ylogy=>
1/xlog(1/x)=1/ylog(1/y)=>
1/y=eW(k,1/xlog(1/x)=>
y=1/eW(k,1/xlog(1/x)
1.Plot function ..k=-1
2.Plot function ..k=0
3.Plot function ..k=1
Differential…
Answer:
1 / 2 √ x - 1
Step-by-step explanation:
Given :
√ x = y + x
Rewrite as :
= > y = √ x - x
Diff. w.r.t. x we get :
= > d y / d x = ( √ x )' - ( x )'
= > d y / d x = 1 / 2 ( 1 / √ x ) - ( x )'
= > d y / d x = 1 / 2 √ x - 1
Hence we required answer!